Answer
$$\frac{8+\sqrt{8}}{2-\sqrt{2}}=\frac{20+12\sqrt{2}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{8+\sqrt{8}}{2-\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{8+\sqrt{8}}{2-\sqrt{2}}\frac{2+\sqrt{2}}{2+\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{16+8\sqrt{2}+4\sqrt{2}+4}{4+2\sqrt{2}-2\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{20+12\sqrt{2}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 + \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 8 + \sqrt{8}\right) } \cdot \left( 2 + \sqrt{2}\right) = \color{blue}{8} \cdot2+\color{blue}{8} \cdot \sqrt{2}+\color{blue}{ \sqrt{8}} \cdot2+\color{blue}{ \sqrt{8}} \cdot \sqrt{2} = \\ = 16 + 8 \sqrt{2} + 4 \sqrt{2} + 4 $$ Simplify denominator. $$ \color{blue}{ \left( 2- \sqrt{2}\right) } \cdot \left( 2 + \sqrt{2}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot \sqrt{2}\color{blue}{- \sqrt{2}} \cdot2\color{blue}{- \sqrt{2}} \cdot \sqrt{2} = \\ = 4 + 2 \sqrt{2}- 2 \sqrt{2}-2 $$ |
| ③ | Simplify numerator and denominator |
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